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In Corollary 11.12 of Atiyah-Macdonald, it says that in a Noetherian ring every prime ideal has finite height. It seems to come directly from Proposition 11.10, which says if $A$ is a Noetherian local ring, $\dim A \leq d(A)$, where $d(A)$ is the degree of the characteristic polynomial of $A$, hence finite. It seems like from that proposition you can only conclude that any prime ideal in a Noetherian local ring has finite height. I don't see how you can reach Corollary 11.12.

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    $\begingroup$ If you believe in Corollary 11.11, then recall that the height of a prime ideal $\mathfrak p$ equals $\dim A_{\mathfrak p}$ (which is a noetherian local ring). $\endgroup$ – user26857 Jan 5 '20 at 22:51
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    $\begingroup$ How would a chain of ideals establishing infinite height not violate the ascending chain condition? (asks a non-algebraist) $\endgroup$ – Eric Towers Jan 5 '20 at 23:08
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    $\begingroup$ @EricTowers: You don't need an infinite ascending chain to get infinite height, just arbitrarily long finite chains. $\endgroup$ – Eric Wofsey Jan 5 '20 at 23:11
  • $\begingroup$ @EricTowers Also, the height of a prime ideal $\mathfrak{p}$ is the supremum of lengths of chains of prime ideals contained in $\mathfrak{p}$, so the height of a prime ideal has something to do with the descending chain condition, not the ascending chain condition. $\endgroup$ – Emily Williams Jan 6 '20 at 20:59
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One should be careful not to confuse the height of an ideal with the Krull dimension of the ring. A famous example of Nagata shows that a Noetherian ring can indeed have infinite Krull dimension, precisely because it has prime ideals which have arbitrarily large height. See this question for details of Nagata's example.

However, the height of any prime ideal $\mathfrak{p}$ in a Noetherian ring $R$ can still never be infinite. To see this, as the comments mention, one can combine Proposition 11.10 in Atiyah-Macdonald with the fact that $\operatorname{ht}\mathfrak{p}=\dim R_{\mathfrak{p}}$, which follows from their Corollary 3.13, to conclude that $\operatorname{ht}\mathfrak{p}$ is finite.

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